Question

Suppose that the metal cylinder in the last problem has a mass of 0.10 kg and the coefficient of static friction between the surface and the cylinder is 0.12. If the cylinder is 0.20 m from the center of the turntable, what is the maximum speed that the cylinder can move along its circular path without slipping off the turntable?

Answer 1

v = 0.485 V

Step-by-step explanation:

Let the Centripetal force be F

`F = (mv^(2) )/(r)`................(1)

mass, m = 0.10 kg

radius, r = 0.20 m

speed, v = ?

`F = \mu mg`..................(2)

`(mv^(2) )/(r) = \mu mg`

`v^(2) = \mu rg`

`v = √(\mu rg)`

`v = √(0.12 * 0.2 * 9.8)`

`v = 0.485 V`

Answer 2

The speed maximum speed is
`0.49ms^(-1)`

Step-by-step explanation:

The centrifugal force always acts on the cylinder and move away the rotating platform from the rotational axis. so the centripetal force provide by the frictional force:

Therefore

`(mv^(2) )/(r) =u_(s) mg`

coefficient of static friction:
`u_(s) =0.12`

mass of the cylinder:
`m=0.10kg`\

distance of the cylinder from the turntable:
`r=0.20m`

`(mv^(2) )/(r) =u_(s) mg`

cross multiply to find v

`v^(2) =u_(s) rg\\v=\sqrt{u_(s)rg } \\v=√(0.12*0.20*9.80)\\ v=0.49m/s`